The Physics
Hypertextbook
Opus in profectus

# Kinematics & Calculus

## Practice

### practice problem 1

Determine the equations of motion for constant jerk.

#### solution

Jerk is the derivative of acceleration. Undo that process. Integrate jerk to get acceleration. That's the first time I've ever said that. I propose we call it the zeroeth equation of motion for constant jerk. The reason why will be apparent after we finish the next derivation.

 j = da dt da = j dt
 a Δt ⌠⌡ da = ⌠⌡ j dt a0 0
 a − a0 = jΔt a = a0 + jΔt [0]

Acceleration is the derivative of velocity. Integrate acceleration to get velocity. We've done this process before. We called the result the velocity–time relationship or the first equation of motion when acceleration was constant. We should give it a similar name. This is the first equation of motion for constant jerk.

 a = dv dt dv = a dt dv = (a0 + jt) dt
 v Δt ⌠⌡ dv = ⌠⌡ (a0 + jt) dt v0 0
 v − v0 = a0Δt + ½jΔt2 v = v0 + a0Δt + ½jΔt2 [1]

Velocity is the derivative of displacement. Integrate velocity to get displacement. We've done this before too. The resulting displacement–time relationship will be our second equation of motion for constant jerk.

 v = dx dt dx = v dt dx = (v0 + a0Δt + ½jΔt2) dt
 x Δt ⌠⌡ dx = ⌠⌡ (v0 + a0Δt + ½jΔt2) dt x0 0
 x − x0 = v0Δt + ½a0Δt2 + ⅙jΔt3 x = x0 + v0Δt + ½a0Δt2 + ⅙jΔt3 [2]

Please notice something about these equations. When jerk is zero, they all revert back to the equations of motion for constant acceleration. Zero jerk means constant acceleration, so all is right with the world we've created. (I never said constant acceleration was realistic. Constant jerk is equally mythical. In hypertextbook world, however, all things are possible.)

Where do we go next? Should we try to remold the velocity–displacement relationship (the third equation of motion) in our new realm of constant jerk?

 v = v0 + a0Δt + ½jΔt2 [1] + x = x0 + v0Δt + ½a0Δt2 + ⅙jΔt3 [2] = v = ƒ(x) [3]

How about an acceleration–displacement relationship (the fourth equation of motion)?

 a = a0 + jΔt [1] + x = x0 + v0Δt + ½a0Δt2 + ⅙jΔt3 [2] = a = ƒ(x) [3]

I don't even know if these can be worked out algebraically. I doubt it. (Look at that scary cubic equation for displacement. That can't be our friend.) At the moment, I really don't care. I don't know what physical significance the results would have.

This is the kind of problem that distinguishes physicists from mathematicians. A mathematician wouldn't necessarily care about the physical significance and just might thank the physicist for an interesting challenge. A physicist wouldn't necessarily care about the answer unless it turned out to be useful, in which case the physicist would definitely thank the mathematician for being so curious.

### practice problem 2

An object's position is described by the following polynomial for 0 to 10 s.

x = t3 − 15t2 + 54t

Where x is in meters, t is in seconds, and positive is forward. Determine…

1. the object's velocity as a function of time
2. the object's acceleration as a function of time
3. the object's maximum velocity
4. the object's minimum velocity
5. the time when the object was moving backward
6. the times when the object returned to its starting position
7. the object's average velocity
8. the object's average speed

#### solution

Draw it first.

1. Velocity is the first derivative of displacement.
 v = dx = 3t2 − 30t + 54 dt
2. Acceleration is the second derivative of displacement or the first derivative of velocity.
 a = dv = 6t − 30 dt
3. In general, the extreme values of a function occur at the endpoints of the domain or at local extrema (if they exist). Let's check the endpoints first. How fast is the object moving at the beginning and end of the time interval?
 v(0) = 3(0)2 − 30(0) + 54 v(0) = +54 m/s v(10) = 3(10)2 − 30(10) + 54 v(10) = +54 m/s

Local extrema occur where the derivative of a function is zero. The extrema of the velocity function can be found at the places where the acceleration is zero.

 a = 6t − 30 = 0 m/s2 t = 5 s

This is the time where acceleration reversed direction. It corresponds to an inflection point on the graph.

How fast was the object moving at this time?

 v(5) = 3(5)2 − 30(5) + 54 v(5) = −21 m/s

We now have all the info needed to answer this question and the next one. The maximum velocity was +54 m/s. It occurred at 0 s and 10 s.

4. The minimum velocity was −21 m/s. It occurred at 5 s.
5. The direction of motion is determined by the sign of the velocity. Direction is reversed whenever the velocity switches sign. Between two signs lies a zero. Let's find the zeroes.
 v = 3t2 − 30t + 54 v = 3(t 2− 10t + 18)
 t = 10 ± √[102 − 4(18)] 2
 t = (5 ± √7) s t = 2.354… s, 7.645… s

We have all the info we need to answer this question, but it's scattered. Let's make a little table summarizing what we know so far.

quantity start of
problem
direction
reversal
minimum
velocity
direction
reversal
end of
problem
time 0 s 2.3 s 5 s 7.6 s 10 s
velocity +54 m/s 0 m/s −21 m/s 0 m/s +54 m/s
direction forward not moving backward not moving forward

The object was moving backward from 2.3 s to 7.6 s.

6. Set the position equation equal to zero to determine the times when the object was at the starting point (x = 0 m).
 x = t3 − 15t2 + 54t = 0 m x = t(t − 6)(t − 9) = 0 m t = 0 s, 6 s, 9 s

7. Average velocity is displacement divided by time. We know the object started at x = 0 m so that means Δx = x(10 s)

 v̅ = Δx Δt
 v̅ = (10)3 − 15(10)2 + 54(10) 10
 v̅ = +4 m/s
8. Average speed is distance divided by time. Finding distance is not easy. We need to find the distance traveled during the three stages of motion — forward, backward, and forward again — and then add the absolute values of the displacements. Before we can do that, we need to actually find the critical positions.
 x = t3 − 15t2 + 54t x(0) = (0)3 − 15(0)2 + 54(0) x(0) = 0 m x(2.354) = (2.354)3 − 15(2.354)2 + 54(2.354) x(2.354) = +57.040… m x(7.645) = (7.645)3 − 15(7.645)2 + 54(7.645) x(7.645) = −17.040… m x(10) = (10)3 − 15(10)2 + 54(10) x(10) = +40 m

Now combine them.

segment start finish distance
moving forward +00 m +57 m 057 m
moving backward +57 m −17 m 074 m
moving forward −17 m +40 m 057 m
overall     188 m

Finally, divide distance by time.

 v̅ = Δx = 188 m = 18.8 m/s Δt 10 s

### practice problem 3

The graph below shows the acceleration of a hydraulic elevator in a four story school building as a function of time.

The graph begins at t = 0 s when the elevator door closed on the second floor and ends at t = 20 s when the door opened on a different floor. Assume that the positive directions for displacement, velocity, and acceleration are upward. Determine…
1. the maximum speed of the elevator
2. the duration of the brief jerk experienced by the elevator centered on 17.5 s
Sketch the corresponding graphs of…
1. velocity-time
2. displacement-time
Determine…
1. the most likely floor on which the elevator stopped

#### solution

1. A quick glance at the graph shows the elevator accelerating downward over 3 s, then coasting, then accelerating upward for 2 s, coasting again, and then accelerating for a brief burst before stopping. The speed increases, remains constant, decreases, remains constant, and decreases a bit more — all in the down direction. The greatest speed would happen at the end of the first triangular region on the acceleration-time graph. The area under this bit is the change in velocity from its initial value of zero. Therefore…
 Δv = area under a-t graphΔv = area of a triangle Δv = ½bhΔv = ½(3.0 s)(−6.0 m/s2) Δv = −9.0 m/s
2. If an elevator is going to work properly it has to stop on a floor to let people off. Therefore, the velocity of the elevator at the end of the graph should be zero. Since change in velocity equal the area under the curve on an acceleration-time graph, we need the total area of the three triangular segments to add up to zero. We've already determined this change for the first triangular segment. Let's repeat it for the second.
 Δv = area under a-t graphΔv = area of a triangle Δv = ½bhΔv = ½(2.0 s)(+8.0 m/s2) Δv = +8.0 m/s

Add these two areas up and you don't get zero, you get…

 Δv = −9.0 m/s + 8.0 m/s Δv = −1.0 m/s

Therefore, the area under the remaining segment must be +1.0 m/s to compensate. Use the same concepts as before, but this time solve for the base (change in time) of the triangle instead of the area (change in speed).

 Δv = Δv = area under a-t grapharea of a triangle Δv = +1.0 m/s = ½bh½Δt (+4.0 m/s2) Δt = +0.50 s
3. The best way to construct the graphs for the next two questions is systematically — beginning from first principles. The rate of change of displacement is called velocity, the rate of change of velocity is called acceleration, and the rate of change of acceleration is called jerk. Yes, you heard me right — jerk. The straight line segments of the graph we started with correspond to intervals of constant jerk. (If they were curved we'd have non-uniform jerk.) Work backward, integrating the constant value of j to get a, then integrating that to get v, then integrating that to get s (or y in this case since it's a vertical problem).
 j = j = j a = ∫ j dt = a0 + j Δt v = ∫ a dt = v0 + a0 Δt + ½ j Δt2 y = ∫ v dt = y0 + v0 Δt + ½ a0 Δt + ⅙ j Δt3

Apply these equations over and over again. This much computation is best left to a computer. The results are summarized in the table below.

A hydraulic elevator
time (s) jerk (m/s3) acceleration (m/s2) velocity (m/s) position (m)
0.00 0.0 0.0 0.00 0.000
1.00 −0.4 0.0 0.00 0.000
2.50 0.4 −0.6 −0.45 −0.225
4.00 0.0 0.0 −0.90 −1.350
13.00 0.8 0.0 −0.90 −9.450
14.00 −0.8 0.8 −0.50 −10.217
15.00 0.0 0.0 −0.10 −10.450
17.25 1.6 0.0 −0.10 −10.675
17.50 −1.6 0.4 −0.05 −10.696
17.75 0.0 0.0 0.00 −10.700
20.00 0.0 0.0 0.00 −10.700
4. Here's the velocity-time graph for the elevator. The horizontal segments correspond to intervals with no acceleration. The curved segments correspond to intervals with changing acceleration.

5. Here's the displacement-time graph for the elevator. The intervals with acceleration are curved. The intervals without acceleration are straight. The beginning and end of the graph are horizontal since the elevator is stopped.

6. The overall displacement of the elevator was 10.7 m below the second floor. Ceiling heights in a typical residential building are about 3 m. Public buildings like schools tend to have taller floors than homes. My guess is that the ceiling heights in this school are on the order of 5 m. That would mean the elevator stopped in the basement.